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Noise in Eigenvalues plot



Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Problem with plotting eigenvaluesHow to overlay ListPlot on a ContourPlot with correct range?Trying to find intersection of 3 functions graphicallySome glitch in the Plot: Two approaches for plotting give different resultsDEigenvalues with Robin B.C. sign problemHow can I add a custom color function and a custom mesh to a 3D parametric plot?How do I plot $y=8 sin(2 pi / 3)$?Plotting eigenvalues in one plot for three different parametersEigenvalues of a non-Hermitian complex periodic potentialHow to compute eigenvalues of a large symbolic matrix?










2












$begingroup$


I am trying to Plot Eigenvalues of a Hamiltonian, but I am getting noisy plot, which is incorrect. Here is the code.



A1 = 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, -1, 0;
A2 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, -I, 0, 0, I, 0;
A3 = 0, 0, 0, -1, 0, 0, 1, 0, 0, 1, 0, 0, -1, 0, 0, 0;
A4 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, I, 0, 0, -I, 0;
A5 = 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1;
A6 = 0, 0, 0, -I, 0, 0, I, 0, 0, -I, 0, 0, I, 0, 0, 0;
A7 = 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0;
A8 = 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, -1;
H[d_, λ_, β_, m_] :=
a (Sin[x] A1 + Sin[ky] A2) + A3 β +
d A4 + (t Cos[z] + 2 b (2 - Cos[x] - Cos[ky])) A5 + α*
Sin[ky] A6 + λ Sin[z] A7+m*A8;
ky = 0;
a = 1;
b = 1;
t = 1.5;
α = 0.3;
Plot3D[Eigenvalues[H[0.1, 0.5, 0.7, 0]][[4]], x, -π, π, z, 0, 2 π]


Mathematica graphics



Any help will be highly appreciated.










share|improve this question











$endgroup$
















    2












    $begingroup$


    I am trying to Plot Eigenvalues of a Hamiltonian, but I am getting noisy plot, which is incorrect. Here is the code.



    A1 = 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, -1, 0;
    A2 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, -I, 0, 0, I, 0;
    A3 = 0, 0, 0, -1, 0, 0, 1, 0, 0, 1, 0, 0, -1, 0, 0, 0;
    A4 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, I, 0, 0, -I, 0;
    A5 = 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1;
    A6 = 0, 0, 0, -I, 0, 0, I, 0, 0, -I, 0, 0, I, 0, 0, 0;
    A7 = 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0;
    A8 = 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, -1;
    H[d_, λ_, β_, m_] :=
    a (Sin[x] A1 + Sin[ky] A2) + A3 β +
    d A4 + (t Cos[z] + 2 b (2 - Cos[x] - Cos[ky])) A5 + α*
    Sin[ky] A6 + λ Sin[z] A7+m*A8;
    ky = 0;
    a = 1;
    b = 1;
    t = 1.5;
    α = 0.3;
    Plot3D[Eigenvalues[H[0.1, 0.5, 0.7, 0]][[4]], x, -π, π, z, 0, 2 π]


    Mathematica graphics



    Any help will be highly appreciated.










    share|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      I am trying to Plot Eigenvalues of a Hamiltonian, but I am getting noisy plot, which is incorrect. Here is the code.



      A1 = 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, -1, 0;
      A2 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, -I, 0, 0, I, 0;
      A3 = 0, 0, 0, -1, 0, 0, 1, 0, 0, 1, 0, 0, -1, 0, 0, 0;
      A4 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, I, 0, 0, -I, 0;
      A5 = 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1;
      A6 = 0, 0, 0, -I, 0, 0, I, 0, 0, -I, 0, 0, I, 0, 0, 0;
      A7 = 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0;
      A8 = 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, -1;
      H[d_, λ_, β_, m_] :=
      a (Sin[x] A1 + Sin[ky] A2) + A3 β +
      d A4 + (t Cos[z] + 2 b (2 - Cos[x] - Cos[ky])) A5 + α*
      Sin[ky] A6 + λ Sin[z] A7+m*A8;
      ky = 0;
      a = 1;
      b = 1;
      t = 1.5;
      α = 0.3;
      Plot3D[Eigenvalues[H[0.1, 0.5, 0.7, 0]][[4]], x, -π, π, z, 0, 2 π]


      Mathematica graphics



      Any help will be highly appreciated.










      share|improve this question











      $endgroup$




      I am trying to Plot Eigenvalues of a Hamiltonian, but I am getting noisy plot, which is incorrect. Here is the code.



      A1 = 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, -1, 0;
      A2 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, -I, 0, 0, I, 0;
      A3 = 0, 0, 0, -1, 0, 0, 1, 0, 0, 1, 0, 0, -1, 0, 0, 0;
      A4 = 0, -I, 0, 0, I, 0, 0, 0, 0, 0, 0, I, 0, 0, -I, 0;
      A5 = 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1;
      A6 = 0, 0, 0, -I, 0, 0, I, 0, 0, -I, 0, 0, I, 0, 0, 0;
      A7 = 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0;
      A8 = 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, -1;
      H[d_, λ_, β_, m_] :=
      a (Sin[x] A1 + Sin[ky] A2) + A3 β +
      d A4 + (t Cos[z] + 2 b (2 - Cos[x] - Cos[ky])) A5 + α*
      Sin[ky] A6 + λ Sin[z] A7+m*A8;
      ky = 0;
      a = 1;
      b = 1;
      t = 1.5;
      α = 0.3;
      Plot3D[Eigenvalues[H[0.1, 0.5, 0.7, 0]][[4]], x, -π, π, z, 0, 2 π]


      Mathematica graphics



      Any help will be highly appreciated.







      plotting eigenvalues






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 55 mins ago









      Michael E2

      151k12203483




      151k12203483










      asked 1 hour ago









      Hazoor ImranHazoor Imran

      213




      213




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          By default, the eigenvalues are ordered by absolute value. All the eigenvalues of this particular matrix have the same absolute value plus some rounding errors. Thus, it can easily happen, that the fourth eigenvalue is positive or negative, depending on the parameters.



          You can use Max to plot the largest eigenvalue:



          Plot3D[Max@Eigenvalues[H[0.1, 0.5, 0.7, 0.]], x, -Pi, Pi, z, 0, 2 Pi]


          enter image description here



          Alternatively, you may use the "Criteria" suboption of the Method "Arnoldi":



          Plot3D[
          Eigenvalues[
          H[0.1, 0.5, 0.7, 0], -1,
          Method -> "Arnoldi", "Criteria" -> "RealPart"
          ],
          x, - Pi, Pi, z, 0, 2 Pi]





          share|improve this answer









          $endgroup$




















            2












            $begingroup$

            Not sure why you pick the 4th element, but maybe this will help:



            ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. 
            Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0];
            Plot3D[ev4, x, -π, π, z, 0, 2 π]


            enter image description here






            share|improve this answer









            $endgroup$












            • $begingroup$
              Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
              $endgroup$
              – Hazoor Imran
              5 mins ago











            Your Answer








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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            By default, the eigenvalues are ordered by absolute value. All the eigenvalues of this particular matrix have the same absolute value plus some rounding errors. Thus, it can easily happen, that the fourth eigenvalue is positive or negative, depending on the parameters.



            You can use Max to plot the largest eigenvalue:



            Plot3D[Max@Eigenvalues[H[0.1, 0.5, 0.7, 0.]], x, -Pi, Pi, z, 0, 2 Pi]


            enter image description here



            Alternatively, you may use the "Criteria" suboption of the Method "Arnoldi":



            Plot3D[
            Eigenvalues[
            H[0.1, 0.5, 0.7, 0], -1,
            Method -> "Arnoldi", "Criteria" -> "RealPart"
            ],
            x, - Pi, Pi, z, 0, 2 Pi]





            share|improve this answer









            $endgroup$

















              2












              $begingroup$

              By default, the eigenvalues are ordered by absolute value. All the eigenvalues of this particular matrix have the same absolute value plus some rounding errors. Thus, it can easily happen, that the fourth eigenvalue is positive or negative, depending on the parameters.



              You can use Max to plot the largest eigenvalue:



              Plot3D[Max@Eigenvalues[H[0.1, 0.5, 0.7, 0.]], x, -Pi, Pi, z, 0, 2 Pi]


              enter image description here



              Alternatively, you may use the "Criteria" suboption of the Method "Arnoldi":



              Plot3D[
              Eigenvalues[
              H[0.1, 0.5, 0.7, 0], -1,
              Method -> "Arnoldi", "Criteria" -> "RealPart"
              ],
              x, - Pi, Pi, z, 0, 2 Pi]





              share|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                By default, the eigenvalues are ordered by absolute value. All the eigenvalues of this particular matrix have the same absolute value plus some rounding errors. Thus, it can easily happen, that the fourth eigenvalue is positive or negative, depending on the parameters.



                You can use Max to plot the largest eigenvalue:



                Plot3D[Max@Eigenvalues[H[0.1, 0.5, 0.7, 0.]], x, -Pi, Pi, z, 0, 2 Pi]


                enter image description here



                Alternatively, you may use the "Criteria" suboption of the Method "Arnoldi":



                Plot3D[
                Eigenvalues[
                H[0.1, 0.5, 0.7, 0], -1,
                Method -> "Arnoldi", "Criteria" -> "RealPart"
                ],
                x, - Pi, Pi, z, 0, 2 Pi]





                share|improve this answer









                $endgroup$



                By default, the eigenvalues are ordered by absolute value. All the eigenvalues of this particular matrix have the same absolute value plus some rounding errors. Thus, it can easily happen, that the fourth eigenvalue is positive or negative, depending on the parameters.



                You can use Max to plot the largest eigenvalue:



                Plot3D[Max@Eigenvalues[H[0.1, 0.5, 0.7, 0.]], x, -Pi, Pi, z, 0, 2 Pi]


                enter image description here



                Alternatively, you may use the "Criteria" suboption of the Method "Arnoldi":



                Plot3D[
                Eigenvalues[
                H[0.1, 0.5, 0.7, 0], -1,
                Method -> "Arnoldi", "Criteria" -> "RealPart"
                ],
                x, - Pi, Pi, z, 0, 2 Pi]






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 50 mins ago









                Henrik SchumacherHenrik Schumacher

                60.7k585171




                60.7k585171





















                    2












                    $begingroup$

                    Not sure why you pick the 4th element, but maybe this will help:



                    ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. 
                    Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0];
                    Plot3D[ev4, x, -π, π, z, 0, 2 π]


                    enter image description here






                    share|improve this answer









                    $endgroup$












                    • $begingroup$
                      Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
                      $endgroup$
                      – Hazoor Imran
                      5 mins ago















                    2












                    $begingroup$

                    Not sure why you pick the 4th element, but maybe this will help:



                    ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. 
                    Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0];
                    Plot3D[ev4, x, -π, π, z, 0, 2 π]


                    enter image description here






                    share|improve this answer









                    $endgroup$












                    • $begingroup$
                      Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
                      $endgroup$
                      – Hazoor Imran
                      5 mins ago













                    2












                    2








                    2





                    $begingroup$

                    Not sure why you pick the 4th element, but maybe this will help:



                    ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. 
                    Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0];
                    Plot3D[ev4, x, -π, π, z, 0, 2 π]


                    enter image description here






                    share|improve this answer









                    $endgroup$



                    Not sure why you pick the 4th element, but maybe this will help:



                    ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. 
                    Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0];
                    Plot3D[ev4, x, -π, π, z, 0, 2 π]


                    enter image description here







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 49 mins ago









                    Michael E2Michael E2

                    151k12203483




                    151k12203483











                    • $begingroup$
                      Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
                      $endgroup$
                      – Hazoor Imran
                      5 mins ago
















                    • $begingroup$
                      Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
                      $endgroup$
                      – Hazoor Imran
                      5 mins ago















                    $begingroup$
                    Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
                    $endgroup$
                    – Hazoor Imran
                    5 mins ago




                    $begingroup$
                    Thanks @ Michael E2, Is it possible to do this with an equation by the contourplot. Like ev4 = Eigenvalues[H[p, q, r, s]][[4]] /. Thread[p, q, r, s -> 0.1, 0.5, 0.7, 0]; ContourPlot[ev4==-0.5, x, -[Pi], [Pi], z, 0, 2 [Pi]]. In my case this is not working.
                    $endgroup$
                    – Hazoor Imran
                    5 mins ago

















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